Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;

public class R419A {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(); int m = sc.nextInt();
		int[][] a = new int[n][m];
		int[] row = new int[n];
		int[] col = new int[m];
		int sum = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				a[i][j] = sc.nextInt();
				sum += a[i][j];
				if (a[i][j] == 0) {
					row[i] = 1;
					col[j] = 1;
				}
			}
		}
		ArrayList<String> moves = new ArrayList<String>();
		while (sum > 0) {
			//check if you can make any move at all
			boolean flagRow = false;
			boolean flagCol = false;
			int r = 0; int c = 0;
			for (int i = 0; i < n; i++) {
				if (row[i] == 0) {
					flagRow = true;
					r = i;
				}
			}
			for (int j = 0; j < m; j++) {
				if (col[j] == 0) {
					flagCol = true;
					c = j;
				}
			}
			if (!flagRow && !flagCol) {
				moves.add("-1");
				break;
			} else {
				//this means possible to make some move
				if (flagRow && flagCol) {
					if (n > m) {
						//then make move on the col & record the move
						for (int i = 0; i < n; i++) {
							if (--a[i][c] == 0) {
								col[c] = 1;
								row[i] = 1;
							}
							sum -= 1;
						}
						c += 1;
						moves.add("col " + c);
					} else {
						//then make move on the row
						for (int i = 0; i < m; i++) {
							if (--a[r][i] == 0) {
								row[r] = 1;
								col[i] = 1;
							}
							sum -= 1;
						}
						r += 1;
						moves.add("row " + r);
					}
				} else if (flagRow) {
					//means can make move on the row
					for (int i = 0; i < m; i++) {
						if (--a[r][i] == 0) {
							row[r] = 1;
							col[i] = 1;
						}
						sum -= 1;
					}
					r += 1;
					moves.add("row " + r);
				} else {
					//means can make move on the col
					for (int i = 0; i < n; i++) {
						if (--a[i][c] == 0) {
							col[c] = 1;
							row[i] = 1;
						}
						sum -= 1;
					}
					c += 1;
					moves.add("col " + c);
				}
			}
		}
		if (moves.size() != 0 && moves.get(moves.size() - 1) == "-1") {
			System.out.println(-1);
		} else {
			System.out.println(moves.size());
			while (moves.size() > 0)
				System.out.println(moves.remove(0));
		}
	}
}